NTA JEE Main 22nd April 2013 Online

Instructions

For the following questions answer them individually

NTA JEE Main 22nd April 2013 Online - Question 31


The density of 3M solution of sodium chloride is 1.252 g mL$$^{-1}$$. The molality of the solution will be : (molar mass, NaCl = 58.5 g mol$$^{-1}$$)

NTA JEE Main 22nd April 2013 Online - Question 32


The wave number of the first emission line in the Balmer series of H-Spectrum is : (R = Rydberg constant)

NTA JEE Main 22nd April 2013 Online - Question 33


The order of increasing sizes of atomic radii among the elements O, S, Se and As is :

NTA JEE Main 22nd April 2013 Online - Question 34


The solubility order for alkali metal fluoride in water is :

NTA JEE Main 22nd April 2013 Online - Question 35


Bond order normally gives idea of stability of a molecular species. All the molecules viz. $$H_2$$, $$Li_2$$ and $$B_2$$ have the same bond order yet they are not equally stable. Their stability order is :

NTA JEE Main 22nd April 2013 Online - Question 36


Given Reaction Energy Change (in kJ):
Li(s) $$\rightarrow$$ Li(g) : 161
Li(g) $$\rightarrow$$ Li$$^+$$(g) : 520
$$\frac{1}{2}F_2(g) \rightarrow$$ F(g) : 77
F(g) + e$$^-$$ $$\rightarrow$$ F$$^-$$(g) : (Electron gain enthalpy)
Li$$^+$$(g) + F$$^-$$(g) $$\rightarrow$$ LiF(s) : -1047
Li(s) + $$\frac{1}{2}F_2$$(g) $$\rightarrow$$ LiF(s) : -617
Based on data provided, the value of electron gain enthalpy of fluorine would be :

NTA JEE Main 22nd April 2013 Online - Question 37


The reaction X $$\rightarrow$$ Y is an exothermic reaction. Activation energy of the reaction for X into Y is 150 kJ mol$$^{-1}$$. Enthalpy of reaction is 135 kJ mol$$^{-1}$$. The activation energy for the reverse reaction, Y $$\rightarrow$$ X will be :

NTA JEE Main 22nd April 2013 Online - Question 38


Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts $$Hg_2Cl_2$$, $$Cr_2(SO_4)_3$$, $$BaSO_4$$ and $$CrCl_3$$ respectively?

NTA JEE Main 22nd April 2013 Online - Question 39


NaOH is a strong base. What will be pH of $$5.0 \times 10^{-2}$$ M NaOH solution? (log 2 = 0.3)

NTA JEE Main 22nd April 2013 Online - Question 40


Values of dissociation constant, $$K_a$$ are given as follows :
Acid             $$K_a$$
HCN             $$6.2 \times 10^{-10}$$
HF               $$7.2 \times 10^{-4}$$
HNO$$_2$$          $$4.0 \times 10^{-4}$$
Correct order of increasing base strength of $$CN^-$$, $$F^-$$ and $$NO_2^-$$ will be :

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